7
$\begingroup$

I'd like to plot the density of state (DOS) for a specific system, say an s-wave BCS superconductor, the Green function of which is

$$G\left(p,\omega\right)=\dfrac{\omega+\xi}{\omega^{2}-\xi^{2}-\Delta^{2}}$$

as given e.g. in Abrikosov, Gor'kov and Dzyaloshinski book, eq.(34.16), where $\xi\approx v_{F}\left(p-p_{F}\right)$ represents the kinetic energy from the chemical potential $E_{F}=v_{F}p_{F}$,, $p$ is the momentum, $\hbar\omega$ is the energy and $\Delta$ the superconducting gap.

A frequent definition for the DOS is

$$\rho=\dfrac{1}{\pi}\lim_{\delta\rightarrow0}\Im\left\{ \int dp G_{R}\left(\omega+\mathbf{i}\delta\right)\right\} $$

using the retarded Green's function (perhaps with an other proportionality factor and/or sign, but that's not the important point here I guess). The so-called retarded Green's function $G_{R}$ is defined as the Green's function analytic in the upper half complex-plane.

  • How to compute and/or plot the DOS $\rho$ from $G$ ?

  • Subsidiary, how to see the state at zero-energy in a p-wave superconductor ? In that case the Green function is something like $$G=\dfrac{\omega+\dfrac{p^{2}-p_{F}^{2}}{2m}}{\omega^{2}-\left(\dfrac{p^{2}-p_{F}^{2}}{2m}\right)^{2}-\left(\Delta p\right)^{2}}$$

Thanks for any comment which helps improving this question. Any other example than superconductivity treated pedagogically would help also.

A few more details: The above Green function verifies

$$\left[\left(\begin{array}{cc} \mathbf{i}\hbar\partial & \Delta\\ -\Delta^{\dagger} & -\mathbf{i}\hbar\partial \end{array}\right)+\dfrac{\hbar^{2}}{2m}\nabla^{2}+\mu\right]\mathbf{G}\left(x_{1},x_{2}\right)=\delta\left(x_{1}-x_{2}\right)$$ with $$\mathbf{G}\left(x_{1},x_{2}\right)=\dfrac{\mathbf{i}}{\hbar}\left\langle \hat{T}\left(\begin{array}{c} -\Psi\left(x_{1}\right)\\ \tilde{\Psi}^{\dagger}\left(x_{1}\right) \end{array}\right)\otimes\left(\Psi^{\dagger}\left(x_{2}\right)\;\tilde{\Psi}\left(x_{2}\right)\right)\right\rangle =\left(\begin{array}{cc} G\left(x_{1},x_{2}\right) & -F\left(x_{1},x_{2}\right)\\ F^{\dagger}\left(x_{1},x_{2}\right) & G^{\dagger}\left(x_{1},x_{2}\right) \end{array}\right)$$ and $\Psi$ some fermionic operator. One defines $\xi=\dfrac{p^{2}-p_{F}^{2}}{2m}\approx v_{F}\left(p-p_{F}\right)$ in the road, and one goes to the momentum space to obtain $G\left(p,\omega\right)$ above. I did not give the $F$ function, see the answer by Meng-Cheng below which contains the full $\mathbf{G}$ for a real $\Delta$.

$\endgroup$
  • $\begingroup$ This may be me being unfamiliar with condensed matter lingo, but Green's functions are objects for differential equations/operators, not for systems. Is this a two-point correlation function/propagator? $\endgroup$ – ACuriousMind May 15 '15 at 12:53
  • $\begingroup$ @ACuriousMind Sorry for having not been explicit. I put more details on the question. $\endgroup$ – FraSchelle May 18 '15 at 16:05
7
$\begingroup$

You did not include the proper infinitesimal imaginary part for frequency in the first equation for the Green's function, which would have given the correct pole structures. So it is not clear whether this is time-ordered, retarded or advanced. The retarded Green's function is

$G_R(\omega,p)=\dfrac{\omega+\xi}{(\omega+i\delta)^2-\xi^2-\Delta^2}$

You can either obtain this directly, or perform an analytical continuation from Matsubara(imaginary-time) Green's function.

The imaginary part is

$\Im G=-{\pi(\omega+\xi)}\Big(\delta(\omega-\sqrt{\xi^2+\Delta^2})+\delta(\omega+\sqrt{\xi^2+\Delta^2}) \Big)$

Another problem is that in a superconductor the Green's function is really a matrix. The one you have is probably just $\langle c c^\dagger\rangle$. The density of state is given by the imaginary part of the trace of the matrix Green's function. The full Green's function is given by

$\dfrac{\omega+\xi\tau_z+\Delta\tau_x}{\omega^2-\xi^2-\Delta^2}$

Here $\tau$ are Pauli matrices in the Nambu (particle-hole) space.

For $p$-wave superconductors, well, once you write down the Green's function in momentum space, you have assumed translation invariance and a constant superconducting gap everywhere. Zero-energy state only appears at defects, where the order parameter vanishes (like the core of a vortex, or the edge). So you don't see the zero-energy state from this particular Green's function because there are not any. You have to solve BdG equation in real space, or solve the full Gor'kov equation for the Green's function in real space to obtain the zero energy state.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.